Running the following code will print out <code>orld</code>. What is happening here? What exactly does <code>&(p[*(i + j)])</code> do? <pre class="prettyprint"><code>#include <stdio.h> char p[] = "HelloWorld"; int i[] = {2,1,3,5,6}, j = 4; int main() { printf(&(p[*(i + j)])); return 0; } </code></pre>

<pre class="prettyprint"><code>char p[] = "HelloWorld"; int i[] = {2,1,3,5,6}, j = 4; </code></pre> <code>&(p[*(i + j)])</code> is evaluated as below: here <code>i</code> is the <code>base address</code> of <code>array i</code>. Thus <code>i+4</code> will be the <code>address</code> of the <code>fifth element</code> in the <code>array i</code>. <code>*(i+j)</code> will be equal to <code>6</code>. <code>P[6]</code> will be <code>o</code> after <code>W</code>. <code>&(p[*(i + j)])</code> would be equal to <code>&p[6]</code>. Thus in <code>printf</code> you are passing the <code>address of o</code> and the output would be <code>orld</code>.

What exactly does &(p[*(i + j)]) do?

Tags:

c

Running the following code will print out orld. What is happening here? What exactly does &(p[*(i + j)]) do?

#include <stdio.h>
char p[] = "HelloWorld";
int i[] = {2,1,3,5,6}, j = 4;

int main()
{
    printf(&(p[*(i + j)]));
    return 0;
}

853

asked Dec 25 '22 12:12

marktani

1 Answers

char p[] = "HelloWorld";
int i[] = {2,1,3,5,6}, j = 4;

&(p[*(i + j)]) is evaluated as below:

here i is the base address of array i. Thus i+4 will be the address of the fifth element in the array i. *(i+j) will be equal to 6. P[6] will be o after W. &(p[*(i + j)]) would be equal to &p[6]. Thus in printf you are passing the address of o and the output would be orld.